Question

A rubber cord \[10\,{\text{m}}\] long is suspended vertically. How much does it stretch under its own weight? (Density of rubber is \[1500\,{\text{kg/}}{{\text{m}}^3}\], \[Y = 5 \times {10^8}\,{\text{N/}}{{\text{m}}^2}\], \[g = 10\,{\text{m}}/{{\text{s}}^{\text{2}}}\])
A. \[15 \times {10^{ - 4}}\,{\text{m}}\]
B. \[7.5 \times {10^{ - 4}}\,{\text{m}}\]
C. \[12 \times {10^{ - 4}}\,{\text{m}}\]
D. \[25 \times {10^{ - 4}}\,{\text{m}}\]

Answer 1

Hint: Use the formula for Young’s modulus for material of a substance. The force acting on the rubber cord is weight of the cord acting at its centre. Determine the weight of the cord in terms of its length, area and density of rubber. Substitute all these values in the formula for Young’s modulus and determine change in length of the rubber cord.

Formulae used:
The Young’s modulus \[Y\] for the material of a wire is given by
\[Y = \dfrac{{FL}}{{A\Delta L}}\] …… (1)
Here, \[f\] is the force on the wire, \[L\] is the original length of the wire, \[A\] is the cross-sectional area of the wire and \[\Delta L\] is the change in length of the wire.
The density \[\rho \] of an object is
\[\rho = \dfrac{M}{V}\] …… (2)
Here, \[M\] is the mass of the object and \[V\] is the volume of the object.
The volume \[V\] of the wire is given by
\[V = AL\] …… (3)
Here, \[A\] is a cross sectional area of wire and \[L\] is length of the wire.

Complete step by step answer:
We have given that the length of the rubber cord is \[10\,{\text{m}}\] and the rubber cord stretches due to its own weight.
\[L = 10\,{\text{m}}\]

Let us determine the weight of the rubber cord which is the only force acting on the rubber cord.
Rearrange equation (2) for mass of the rubber cord.
\[M = \rho V\]

Let \[M\] be the mass of the rubber cord. Hence, the weight of the rubber cord is
\[W = Mg\]
Substitute \[\rho V\] for \[M\] in the above equation.
\[W = \rho Vg\]
Substitute \[AL\] for \[V\] in the above equation.
\[W = \rho ALg\]

The weight of the cord is acting at its centre in the downward direction. Hence, only the upper half length of the cord is stretched by the weight of the cord.

Hence, the length of the cord being stretched by the weight of rubber cord is \[L = \dfrac{{10\,{\text{m}}}}{2} = 5\,{\text{m}}\].
Substitute \[W\] for \[F\] in equation (1).
\[Y = \dfrac{{WL}}{{A\Delta L}}\]

Substitute \[\rho ALg\] for \[W\] in the above equation and rearrange it for \[\Delta L\].
\[Y = \dfrac{{\rho ALgL}}{{A\Delta L}}\]
\[ \Rightarrow Y = \dfrac{{\rho {L^2}g}}{{\Delta L}}\]
\[ \Rightarrow \Delta L = \dfrac{{\rho {L^2}g}}{Y}\]

Substitute \[1500\,{\text{kg/}}{{\text{m}}^3}\] for \[\rho \], \[5\,{\text{m}}\] for \[L\], \[10\,{\text{m}}/{{\text{s}}^{\text{2}}}\] for \[g\] and \[5 \times {10^8}\,{\text{N/}}{{\text{m}}^2}\] for \[Y\] in the above equation.
\[ \Rightarrow \Delta L = \dfrac{{\left( {1500\,{\text{kg/}}{{\text{m}}^3}} \right){{\left( {5\,{\text{m}}} \right)}^2}\left( {10\,{\text{m}}/{{\text{s}}^{\text{2}}}} \right)}}{{5 \times {{10}^8}\,{\text{N/}}{{\text{m}}^2}}}\]
\[ \Rightarrow \Delta L = 7.5 \times {10^{ - 4}}\,{\text{m}}\]
\[ \therefore \Delta L = 15 \times {10^{ - 5}}\,{\text{m}}\]
Therefore, the stretching in the rubber cord due to its own weight is \[7.5 \times {10^{ - 4}}\,{\text{m}}\].

Hence, the correct option is B.

Note:The students should keep in mind that the weight of the rubber cord is acting at its centre. Hence, only the upper half of its length stretches. Therefore, the length of the wire used in the formula for Young’s modulus is half of its original length. Otherwise, the final answer will not be correct.